Answer:
Step-by-step explanation:
This is a test of 2 independent groups. The population standard deviations are not known. Let μ1 be the mean compression strength from the extra carbonation of strawberry drink and μ2 be the mean compression strength from the extra carbonation of cola.
The random variable is μ1 - μ2 = difference in the mean compression strength from the extra carbonation of strawberry drink and the mean compression strength from the extra carbonation of cola.
We would set up the hypothesis.
The null hypothesis is
H0 : μ1 = μ2 H0 : μ1 - μ2 = 0
The alternative hypothesis is
H1 : μ1 < μ2 H1 : μ1 - μ2 < 0
This is a left tailed test.
Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is
(x1 - x2)/√(s1²/n1 + s2²/n2)
From the information given,
x1 = 530
x2 = 553
s1 = 23
s2 = 16
n1 = 10
n2 = 10
t = (530 - 533)/√(23²/10 + 16²/10)
t = - 2.6
The formula for determining the degree of freedom is
df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²
df = [23²/10 + 16²/10]²/[(1/10 - 1)(23²/10)² + (1/10 - 1)(16²/10)²] = 6162.25/383.752
df = 16
We would determine the probability value from the t test calculator. It becomes
p value = 0.0097
Since alpha, 0.05 > than the p value, 0.0097, then we would reject the null hypothesis. Therefore, at 5% significance level, the data suggests that the extra carbonation of cola results in a higher average compression strength.
and inner radius 
![\displaystyle V = \int_0^1 \pi\Big[ \left( 2 - x^2 \right)^2 - \left( 2 - \sqrt{x}\right)^2\Big] dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20V%20%3D%20%5Cint_0%5E1%20%5Cpi%5CBig%5B%20%5Cleft%28%202%20-%20x%5E2%20%5Cright%29%5E2%20-%20%5Cleft%28%202%20-%20%5Csqrt%7Bx%7D%5Cright%29%5E2%5CBig%5D%20dx)
